Question

The set of all real values of $\lambda$ for which the function $f(x)=\left(1-{\cos }^{2}x\right)\cdot (\lambda +\sin x)$, $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, has exactly one maxima and exactly one minima, is :

• A. $\left(-\frac{1}{2},\frac{1}{2}\right)-\{0\}$
• B. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• C. $\left(-\frac{3}{2},\frac{3}{2}\right)$
• D. $\left(-\frac{3}{2},\frac{3}{2}\right)-\{0\}$

Answer 1

Answer: (D)

$f(x)=\left(1-{\cos }^{2}x\right)(\lambda +\sin x)$
$x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$
$f(x)=\lambda {\sin }^{2}x+{\sin }^{3}x$
$f^{\prime }(x)=2\lambda \sin x\cos x+3{\sin }^{2}x\cos x$
$f^{\prime }(x)=\sin x\cos x(2\lambda +3\sin x)$
$\sin x=0,\frac{-2\lambda }{3},(\lambda \ne 0)$
for exactly one maxima & minima
$\frac{-2\lambda }{3}\in (-1,1)\Rightarrow \lambda \in \left(\frac{-3}{2},\frac{3}{2}\right)$
$\lambda \in \left(-\frac{3}{2},\frac{3}{2}\right)-\{0\}$